Let $g(x)=-2\cos(x)+3x^3$. $g'(x)=$
The expression for $g(x)$ includes $\cos(x)$. Remember that the derivative of $\cos(x)$ is $-\sin(x)$. Put another way, $\dfrac{d}{dx}[\cos(x)]=-\sin(x)$. $\begin{aligned} g'(x)&=\dfrac{d}{dx}[-2\cos(x)+3x^3] \\\\ &=-2\dfrac{d}{dx}[\cos(x)]+3\dfrac{d}{dx}(x^3) \\\\ &=-2(-\sin(x))+3\cdot3x^2 \\\\ &=2\sin(x)+9x^2 \end{aligned}$ In conclusion, $g'(x)=2\sin(x)+9x^2$